OFFSET
3,2
COMMENTS
See A073101 for more details.
a(n) = floor(n/4) + 1, at least up to n = 2000, except for some n = 8k+1 (k = 6, 9, 11, 14, 20, 21, 24, 29, 30, 35, 39, 41, 44, 45, 50, ...), where a(n) is one larger than a(n-1) and a(n+1). - M. F. Hasler, Jul 02 2022
LINKS
M. F. Hasler, Table of n, a(n) for n = 3..2000
FORMULA
Conjecture: a(n) = floor(n/4) + d, with d = 1 except for some n = 8k+1 (k = 6, 9, 11, 14, 20, 21, 24, 29, 30, 35, 39, ...) where d = 2 . - M. F. Hasler, Jul 02 2022
EXAMPLE
For n = 3, we have a(3) = 1 = x in 4/3 = 1/x + 1/y + 1/z with y = 4 and z = 12 which is the largest possible z: Indeed, x < y < z gives 4/3 < 3/x, so only x = 1 and 2 are possible, and then with y < z, 2/y > 4/3 - 1/x is impossible for x = 2 < y < 12/5 and for x = 1 < y < 6 only y = 4 gives a solution.
MAPLE
A075245:= proc () local t, n, a, b, t1, largex, largez; for n from 3 to 100 do t := 4/n; largez := 0; largex := 0; for a from floor(1/t)+1 to floor(3/t) do t1 := t-1/a; for b from max(a, floor(1/t1)+1) to floor(2/t1) do if `and`(type(1/(t1-1/b), integer), a < b, b < 1/(t1-1/b)) then if largez < 1/(t1-1/b) then largez := 1/(t1-1/b); largex := a end if end if end do end do; lprint(n, largex) end do end proc; # [program derived from A192787] Patrick J. McNab, Aug 20 2014
MATHEMATICA
For[xLst={}; yLst={}; zLst={}; n=3, n<=100, n++, cnt=0; xr=n/4; If[IntegerQ[xr], x=xr+1, x=Ceiling[xr]]; While[yr=1/(4/n-1/x); If[IntegerQ[yr], y=yr+1, y=Ceiling[yr]]; cnt==0&&y>x, While[zr=1/(4/n-1/x-1/y); cnt==0&&zr>y, If[IntegerQ[zr], z=zr; cnt++; AppendTo[xLst, x]; AppendTo[yLst, y]; AppendTo[zLst, z]]; y++ ]; x++ ]]; xLst
PROG
(PARI) apply( {A075245(n, c=1, a)=for(x=n\4+1, 3*n\4, my(t=4/n-1/x); for(y=max(1\t, x)+1, ceil(2/t)-1, t-1/y >= c && break; numerator(t-1/y)==1 && [c, a]=[t-1/y, x])); a}, [3..99]) \\ M. F. Hasler, Jul 02 2022
CROSSREFS
KEYWORD
hard,nice,nonn
AUTHOR
T. D. Noe, Sep 10 2002
STATUS
approved